\(\text{Đk:}x\ne-\frac{1}{2}\Rightarrow P=\frac{4x^2\left(x+2\right)-\left(x+2\right)}{\left(2x+1\right)^2}=\frac{\left(4x^2-1\right)\left(x+2\right)}{\left(2x+1\right)^2}=\frac{\left(2x-1\right)\left(x+2\right)}{2x+1}\)

\(=\frac{2x^2+4x-x-2}{2x+1}=\frac{3}{2}\Rightarrow2x^2+3x-2=3x+\frac{3}{2}\Leftrightarrow2x^2-\frac{7}{2}=0......\)

\(P\text{ nguyên }\Rightarrow2x^2+3x-2⋮2x+1\Leftrightarrow2x^2+3x-2-\left(x+1\right)\left(2x+1\right)⋮2x+1\Leftrightarrow-3⋮2x+1....\)

a) \(P=\frac{4x^3+8x^2+x-2}{4x^2+4x+1}=\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}\)

ĐKXĐ :\(\left(2x+1\right)^2\ne0=>2x+1\ne0=>x\ne-\frac{1}{2}\)

b) \(P=\frac{3}{2}\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}=\frac{3}{2}\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)}{2x+1}=\frac{3}{2}\Leftrightarrow4x^2-2x+8x-4=6x+3\)

\(\Rightarrow4x^2=7=>x^2=\frac{7}{4}=>x=\pm\sqrt{\frac{7}{4}}\)

c) \(P=\frac{\left(x+2\right)\left(2x-1\right)}{\left(2x+1\right)}=\frac{\left(x+2\right)\left(2x+1-2\right)}{2x+1}=\frac{\left(x+2\right)\left(2x+1\right)-2\left(x+2\right)}{2x+1}\)

\(=x+2-\frac{2x+2}{2x+1}=x+2-1-\frac{1}{2x+1}\)

để P nguyền khi zà chỉ khi

\(1⋮2x+1\)

\(=>2x+1\inƯ\left(1\right)=\pm1\)

=>\(\orbr{\begin{cases}2x+1=1\\2x+1=-1\end{cases}=>\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

a) đk: \(x\ne-\frac{1}{2}\)

b) \(P=\frac{3}{2}\Leftrightarrow\frac{4x^3+4x^2-x-2}{4x^2+4x+1}=\frac{3}{2}\)

\(\Leftrightarrow8x^3+8x^2-2x-4=12x^2+12x+3\)

\(\Leftrightarrow8x^3-4x^2-14x-7=0\)

Cardano ra

c) \(P=\frac{4x^3+4x^2-x-2}{4x^2+4x+1}=x-\frac{2x+2}{4x^2+4x+1}\)

Xét delta tìm khoảng giá trị của biến P

mk cần gấp lắm các bạn ạk

BÀI 1:

a)  \(ĐKXĐ:\)          \(x-3\)\(\ne\)\(0\)

                          \(\Leftrightarrow\)\(x\)\(\ne\)\(3\)

b)   \(A=\frac{x^3-3x^2+4x-1}{x-3}\)

\(=\frac{\left(x^3-3x^2\right)+\left(4x-12\right)+11}{x-3}\)

\(=\frac{x^2\left(x-3\right)+4\left(x-3\right)+11}{x-3}\)

\(=x^2+4+\frac{11}{x-3}\)

Để  \(A\)có giá trị nguyên thì  \(\frac{11}{x-3}\)có giá trị nguyên

hay  \(x-3\)\(\notinƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

Ta lập bảng sau

\(x-3\)    \(-11\)         \(-1\)             \(1\)           \(11\)

\(x\)             \(-8\)               \(2\)              \(4\)           \(14\)

Vậy....

cảm ơn bạn nha nhưng bạn có chắc là nó đúng ko

[2x-2=0=>x=1

x-1=0=>x=1

x+1=0=>x=-1

5=0=>x=5

1.a)\(\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)

\(=\frac{x^3}{\left(x+2\right)\left(x-2\right)}-\frac{x}{x-2}-\frac{2}{x+2}\)

Để biểu thức được xác định thì:\(\left(x+2\right)\left(x-2\right)\ne0\)\(\Rightarrow x\ne\pm2\)

                                                      \(\left(x+2\right)\ne0\Rightarrow x\ne-2\)

                                                      \(\left(x-2\right)\ne0\Rightarrow x\ne2\)

                         Vậy để biểu thức xác định thì : \(x\ne\pm2\)

b) để C=0 thì ....

1, c , bn Nguyễn Hữu Triết chưa lm xong 

ta có : \(/x-5/=2\)

\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}\)

thay x = 7  vào biểu thứcC

\(\Rightarrow C=\frac{4.7^2\left(2-7\right)}{\left(7-3\right)\left(2+7\right)}=\frac{-988}{36}=\frac{-247}{9}\)KL :>...

thay x = 3 vào C 

\(\Rightarrow C=\frac{4.3^2\left(2-3\right)}{\left(3-3\right)\left(3+7\right)}\)

=> ko tìm đc giá trị C tại x = 3

chết mk nhìn nhầm phần c bài 2 :

\(2,\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^2-3x}{2x^2-x^3}\)

Để P xác định 

\(\Rightarrow2-x\ne0\Rightarrow x\ne2\)

\(2+x\ne0\Rightarrow x\ne-2\)

\(x^2-4\ne0\Rightarrow x\ne0\)

\(x^2-3x\ne0\Rightarrow x\ne3\)

b, \(P=\left(\frac{2+x}{2-x}+\frac{4x^2}{\left(2+x\right)\left(2-x\right)}+\frac{2-x}{2+x}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(P=\left[\frac{4+4x+x^2}{\left(2-x\right)\left(2+x\right)}-\frac{4x^2}{\left(2+x\right)\left(2-x\right)}-\frac{4-4x+x^2}{\left(2+x\right)\left(2-x\right)}\right].\frac{x\left(2-x\right)}{x-3}\)

\(P=\left[\frac{8x-4x^2}{\left(2-x\right)\left(2+x\right)}\right].\frac{x\left(2-x\right)}{x-3}=\frac{4x\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)

\(P=\frac{4x^2\left(2-x\right)}{\left(x-3\right)\left(2+x\right)}\)

d, ĐỂ \(p=\frac{8x^2-4x^3}{x^2-x-6}< 0\)

\(TH1:8x^2-4x^3< 0\)

\(\Rightarrow8x^2< 4x^3\)

\(\Rightarrow2< x\Rightarrow x>2\)

\(TH2:x^2-x-6< 0\Rightarrow x^2< x+6\)

a) ĐKXĐ: \(\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\\2-x\ne0\end{cases}}\) => \(\hept{\begin{cases}x\ne-2\\x\ne\pm2\\x\ne2\end{cases}}\) => \(x\ne\pm2\)

Ta có:Q = \(\frac{x-1}{x+2}+\frac{4x+4}{x^2-4}+\frac{3}{2-x}\)

Q = \(\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{4x+4}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

Q = \(\frac{x^2-2x-x+2+4x+4-3x-6}{\left(x+2\right)\left(x-2\right)}\)

Q = \(\frac{x^2-2x}{\left(x+2\right)\left(x-2\right)}=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{x}{x+2}\)

b) ĐKXĐ P: x - 3 \(\ne\)0 => x \(\ne\)3

Ta có: P = 3 => \(\frac{x+2}{x-3}=3\)

=> x + 2 = 3(x - 3)

=> x + 2 = 3x - 9

=> x - 3x = -9 - 2

=> -2x = -11

=> x = 11/2 (tm)

Với x = 11/2 thay vào Q => Q = \(\frac{\frac{11}{2}}{\frac{11}{2}+2}=\frac{11}{15}\)

c) Với x \(\ne\)\(\pm\)2; x \(\ne\)3

Ta có: M = PQ = \(\frac{x+2}{x-3}\cdot\frac{x}{x+2}=\frac{x}{x-3}=\frac{x-3+3}{x-3}=1+\frac{3}{x-3}\)

Để M \(\in\)Z <=> 3 \(⋮\)x - 3

=> x - 3 \(\in\)Ư(3) = {1; -1; 3; -3}

Lập bảng:

Vậy ...

a, ĐKXĐ: x\(\ne\) 1;-1;2

b, A= \(\left(\frac{x}{x+1}+\frac{1}{x-1}-\frac{4x}{2-2x^2}\right):\frac{x+1}{x-2}\)

=\(\left(\frac{2x^2-2x}{2\left(x+1\right)\left(x-1\right)}+\frac{2x+2}{2\left(x+1\right)\left(x-1\right)}+\frac{4x}{2\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-2}{x+1}\)

=\(\frac{2x^2-2x+2x+2+4x}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)

=\(\frac{2x^2+4x+2}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)

=\(\frac{2\left(x+1\right)^2}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)

=\(\frac{x-2}{x-1}\)

c, Khi x= -1

→A= \(\frac{-1-2}{-1-1}\)

= -3

Vậy khi x= -1 thì A= -3

Câu d thì mình đang suy nghĩ nhé, mình sẽ quay lại trả lời sau ^^

a,ĐKXĐ:x#1; x#-1; x#2

b,Ta có:

A=\(\left(\frac{x}{x+1}+\frac{1}{x-1}-\frac{4x}{2-2x^2}\right):\frac{x+1}{x-2}\)

=\(\left(\frac{x\left(x-1\right)2}{\left(x+1\right)\left(x-1\right)2}+\frac{\left(x+1\right)2}{\left(x-1\right)\left(x+1\right)2}+\frac{4x}{2\left(x-1\right)\left(x+1\right)}\right):\frac{x+1}{x-2}\)

=\(\frac{2x^2-2x+2x+2+4x}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)

=\(\frac{2x^2+4x+2}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)

=\(\frac{2\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)

=\(\frac{x-2}{x+1}\)

c,Tại x=-1 ,theo ĐKXĐ x#-1 \(\Rightarrow\)A không có kết quả

d,Để A có giá trị nguyên \(\Rightarrow\frac{x-2}{x+1}\)có giá trị nguyên

\(\Leftrightarrow x-2⋮x+1\)

\(\Leftrightarrow x+1-3⋮x+1\)

Mà \(x+1⋮x+1\Rightarrow3⋮x+1\)

\(\Rightarrow x+1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow x\in\left\{0;-2;2;-4\right\}\)

Mà theo ĐKXĐ x#2\(\Rightarrow x\in\left\{0;-2;-4\right\}\)

Vậy \(x\in\left\{0;-2;-4\right\}\)thì a là số nguyên